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# Nature Of Electromagnetic Waves - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Nature of Electromagnetic Waves is considered one the most difficult concept.

• 92 Questions around this concept.

## Solve by difficulty

Electromagnetic waves are transverse in nature is evident by

The electric field and magnetic field components of an electromagnetic wave going through the vacuum are described by

Then the correct relation between is given by :

In an electromagnetic wave, at an instant and at a particular position, the electric field is along the negative z-axis and the magnetic field is along the positive x-axis. Then the direction of propagation of the electromagnetic wave is :

## Concepts Covered - 1

Nature of Electromagnetic Waves

Nature of Electromagnetic Waves

From Maxwell’s equations, we can observe that electric and magnetic fields in an electromagnetic wave are perpendicular to each other, and to the direction of propagation. Also from our discussion of the displacement current, in that capacitor, the electric field inside the plates of the capacitor is directed perpendicular to the plates. The figure given below shows a typical example of a plane electromagnetic wave propagating along the z direction (the fields are shown as a function of the z coordinate, at a given time t). The electric field Ex is along the x-axis, and varies sinusoidally with z, at a given time. The magnetic field By is along the y-axis, and again varies sinusoidally with z.The electric and magnetic fields Ex and By are perpendicular to each other, and to the direction z of propagation.

Now from the Lorentz equation -

\begin{array}{l}{\text { }} \\ {\qquad \begin{aligned} \vec{F}=& q(\vec{E}+\vec{v} \times \vec{B}) \\ \\ E_{z}=E z_{0} \sin (\omega t-k y) & \\ \\ B_{x}=B x_{0} \sin (\omega t-k y), & \text { where } \frac{\omega}{k}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \end{aligned}}\end{array}

since,  $\omega=2 \pi f$, where f is the frequency and    $k= \frac{2 \pi }{\lambda}$,   where $\lambda$ is the wavelength.

$\text { Therefore, } \frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=f \lambda$

But $f \lambda$ gives the velocity of the wave. So, $f \lambda = c = \omega k$. So we can write -

$c=\frac{\omega}{k}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$

It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as -

$B_{0}= \frac{E_o}{c}$

In a material medium of permittivity ε and magnetic permeability µ, the velocity of light becomes,

$v=\frac{1}{\sqrt{\mu \varepsilon}}$

## Study it with Videos

Nature of Electromagnetic Waves

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